数列an满足a1=1/2,nan+1(下标)-(n+1)an=n^2+n,求数列an/n的前n项和n+1是下标
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数列an满足a1=1/2,nan+1(下标)-(n+1)an=n^2+n,求数列an/n的前n项和n+1是下标
数列an满足a1=1/2,nan+1(下标)-(n+1)an=n^2+n,求数列an/n的前n项和
n+1是下标
数列an满足a1=1/2,nan+1(下标)-(n+1)an=n^2+n,求数列an/n的前n项和n+1是下标
在关系式
na(n+1)-(n+1)an=n(n+1)中,两边都除以n(n+1),得到
a(n+1)/[n+1]-an/n=1
于是可知数列{an/n}是首项为1/2的公差为1的等差数列.
于是an/n=0.5+(n-1)=n-0.5
于是其前n项和为n*(n-0.5+0.5)/2=0.5n^2
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